well that first statement is from the epsilon-delta definition of a limit. I know they don't equal each other, but they do have a relationship, and that's how you approach the problem, that's what I think.

Staff: Mentor

You need to show that for any positive epsilon, there is a positive number delta such that if |(x, y) - (a, b)| < delta, then |x^2 + y^2 - a^2 - b^2)| < epsilon.

What the first inequality says is that (x, y) can be any point within delta of the point (a, b). I'm having a hard time following your work, especially the part just before you said you "opened" the right side.

So a method you could use is to start with [tex](x-a)^2 + (y-b)^2 < \delta^2[/tex], and use it to show that it implies some other inequality [tex](x-a)(x+a) + (y-b)(y+b) < f(\delta)[/tex]. Then if you can show that f is invertible, so that for every ε>0 there exists δ>0 so that f(δ) = ε, then we have found a way to satisfy our definition. The first step in this method is to use the starting inequality and get some basic, weaker inequalities from it, so do that first and play around with it.

With functions of two variables, I find it easiest to "translate" so that the limit is as (x, y) goes to (0, 0) and then change to polar coordinates. That way the distance to (0, 0) is measured by the single variable, r.

That will have limit, as (x, y) goes to (a, b), [itex]a^2+ b^2[/itex] if and only if [itex]r^2+ 2r(a cos(\theta)+ b sin(\theta)= r(r+ 2(acos(\theta)+ b sin(\theta))[/itex] goes to 0 as r goes to 0 (for any value of [itex]\theta[/itex]).